C++析构函数的奇怪行为

在调试器中运行会将内存分配库更改为执行更多检查的内存分配库.除了内存分配和解除分配之外什么都不做的程序将比"正常"程序遭受更多的痛苦.

编辑 刚刚尝试在VS下运行程序我得到一个看起来像的调用堆栈

ntdll.dll!_RtlpValidateHeapEntry@12()  + 0x117 bytes    
ntdll.dll!_RtlDebugFreeHeap@12()  + 0x97 bytes  
ntdll.dll!_RtlFreeHeapSlowly@12()  + 0x228bf bytes  
ntdll.dll!_RtlFreeHeap@12()  + 0x17646 bytes    
msvcr90d.dll!_free_base(void * pBlock=0x0061f6e8)  Line 109 + 0x13 bytes
msvcr90d.dll!_free_dbg_nolock(void * pUserData=0x0061f708, int nBlockUse=1)
msvcr90d.dll!_free_dbg(void * pUserData=0x0061f708, int nBlockUse=1) 
msvcr90d.dll!operator delete(void * pUserData=0x0061f708)
desc.exe!std::allocator::deallocate(int * _Ptr=0x0061f708, unsigned int __formal=4)
desc.exe!std::vector >::_Tidy()  Line 1134  C++

其中显示了ntdll.dll中的调试功能以及正在使用的C运行时.

在调试器中运行会将内存分配库更改为执行更多检查的内存分配库.除了内存分配和解除分配之外什么都不做的程序将比"正常"程序遭受更多的痛苦.

编辑 刚刚尝试在VS下运行程序我得到一个看起来像的调用堆栈

ntdll.dll!_RtlpValidateHeapEntry@12()  + 0x117 bytes    
ntdll.dll!_RtlDebugFreeHeap@12()  + 0x97 bytes  
ntdll.dll!_RtlFreeHeapSlowly@12()  + 0x228bf bytes  
ntdll.dll!_RtlFreeHeap@12()  + 0x17646 bytes    
msvcr90d.dll!_free_base(void * pBlock=0x0061f6e8)  Line 109 + 0x13 bytes
msvcr90d.dll!_free_dbg_nolock(void * pUserData=0x0061f708, int nBlockUse=1)
msvcr90d.dll!_free_dbg(void * pUserData=0x0061f708, int nBlockUse=1) 
msvcr90d.dll!operator delete(void * pUserData=0x0061f708)
desc.exe!std::allocator::deallocate(int * _Ptr=0x0061f708, unsigned int __formal=4)
desc.exe!std::vector >::_Tidy()  Line 1134  C++

其中显示了ntdll.dll中的调试功能以及正在使用的C运行时.